Tex*_*Tex 4 java language-agnostic nlp
I have implemented Hobbs' algorithm for anaphora resolution together with Lappin & Leass ranking for alternatives.
What bugs me is that the description of the algorithm is completely informal, and since there are sentences that are not correctly resolved by my implementation I am not sure whether the limit is on my implementation or on the actual algorithm.
Here is the version I have worked with, found in the Jurafsky&Martin:
- Begin at the noun phrase (NP) node immediately dominating the pronoun.
- Go up the tree to the first NP or sentence (S) node encountered. Call this node X, and call the path used to reach it p.
- Traverse all branches below node X to the left of path p in a left-to-right, breadth-first fashion. Propose as the antecedent any NP node that is encountered which has an NP or S node between it and X.
- If node X is the highest S node in the sentence, traverse the surface parse trees of previous sentences in the text in order of recency, the most recent first; each tree is traversed in a left-to-right, breadth-first manner, and when an NP node is encountered, it is proposed as antecedent. If X is not the highest S node in the sentence, continue to step 5.
- From node X, go up the tree to the first NP or S node encountered. Call this new node X, and call the path traversed to reach it p.
- If X is an NP node and if the path p to X did not pass through the Nominal node that X immediately dominates, propose X as the antecedent.
- Traverse all branches below node X to the left of path p in a left-to-right, breadth- first manner. Propose any NP node encountered as the antecedent.
- If X is an S node, traverse all branches of node X to the right of path p in a left-to- right, breadth-first manner, but do not go below any NP or S node encountered. Propose any NP node encountered as the antecedent.
- Go to Step4
Look at step 3: "to the left of path p". The way I interpreted it is to iterate through the subtrees left-to-right, until I find the branch that contains the pronoun (hence part of the path from the pronoun to X). In Java:
for (Tree relative : X.children()) {
for (Tree candidate : relative) {
if (candidate.contains(pronoun)) break; // I am looking to all the nodes to the LEFT (i.e. coming before) the path leading to X. contain <-> in the path
...
However, doing it this way does not process sentences like "The house is of King Arthur himself". This is due to the fact that "King Arthur" contains "himself" and is therefore not taken into account.
这是Hobbs算法的限制还是我在这里误会了什么？
作为参考,Java中的完整代码(使用Stanford Parser)在这里:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.Reader;
import java.io.StringReader;
import java.io.StringWriter;
import java.util.ArrayList;
import java.util.Collection;
import java.util.List;
import java.util.Set;
import java.util.StringTokenizer;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerConfigurationException;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import org.apache.commons.lang3.ArrayUtils;
import org.apache.commons.lang3.StringUtils;
import org.apache.commons.lang3.StringEscapeUtils;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.NamedNodeMap;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;
import edu.stanford.nlp.ling.HasWord;
import edu.stanford.nlp.ling.Word;
import edu.stanford.nlp.ling.Sentence;
import edu.stanford.nlp.process.DocumentPreprocessor;
import edu.stanford.nlp.process.Tokenizer;
import edu.stanford.nlp.trees.*;
import edu.stanford.nlp.parser.lexparser.LexicalizedParser;
class ParseAllXMLDocuments {
/**
* @throws ParserConfigurationException
* @throws SAXException
* @throws TransformerException
* @throws ModifyException
* @throws NavException
* @throws TranscodeException
* @throws ParseException
* @throws EntityException
* @throws EOFException
* @throws EncodingException */
static final int MAXPREVSENTENCES = 4;
public static void main(String[] args) throws IOException, SAXException, ParserConfigurationException, TransformerException {
// File dataFolder = new File("DataToPort");
// File[] documents;
String grammar = "grammar/englishPCFG.ser.gz";
String[] options = { "-maxLength", "100", "-retainTmpSubcategories" };
LexicalizedParser lp =
new LexicalizedParser(grammar, options);
//
// if (dataFolder.isDirectory()) {
// documents = dataFolder.listFiles();
// } else {
// documents = new File[] {dataFolder};
// }
// int currfile = 0;
// int totfiles = documents.length;
// for (File paper : documents) {
// currfile++;
// if (paper.getName().equals(".DS_Store")||paper.getName().equals(".xml")) {
// currfile--;
// totfiles--;
// continue;
// }
// System.out.println("Working on "+paper.getName()+" (file "+currfile+" out of "+totfiles+").");
//
// DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance(); // This is for XML
// DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
// Document doc = docBuilder.parse(paper.getAbsolutePath());
//
// NodeList textlist = doc.getElementsByTagName("text");
// for(int i=0; i < textlist.getLength(); i++) {
// Node currentnode = textlist.item(i);
// String wholetext = textlist.item(i).getTextContent();
String wholetext = "The house of King Arthur himself. You live in it all the day.";
//System.out.println(wholetext);
//Iterable<List<? extends HasWord>> sentences;
System.out.println(wholetext);
ArrayList<Tree> parseTrees = new ArrayList<Tree>();
String asd = "";
int j = 0;
StringReader stringreader = new StringReader(wholetext);
DocumentPreprocessor dp = new DocumentPreprocessor(stringreader);
@SuppressWarnings("rawtypes")
ArrayList<List> sentences = preprocess(dp);
for (List sentence : sentences) {
parseTrees.add( lp.apply(sentence) ); // Parsing a new sentence and adding it to the parsed tree
ArrayList<Tree> PronounsList = findPronouns(parseTrees.get(j)); // Locating all pronouns to resolve in the sentence
Tree corefedTree;
for (Tree pronounTree : PronounsList) {
parseTrees.set(parseTrees.size()-1, HobbsResolve(pronounTree, parseTrees)); // Resolving the coref and modifying the tree for each pronoun
}
StringWriter strwr = new StringWriter();
PrintWriter prwr = new PrintWriter(strwr);
TreePrint tp = new TreePrint("penn");
tp.printTree(parseTrees.get(j), prwr);
prwr.flush();
asd += strwr.toString();
j++;
}
String armando = "";
for (Tree sentence : parseTrees) {
for (Tree leaf : Trees.leaves(sentence))
armando += leaf + " ";
}
System.out.println(armando);
System.out.println("All done.");
// currentnode.setTextContent(asd);
// }
// TransformerFactory transformerFactory = TransformerFactory.newInstance();
// Transformer transformer = transformerFactory.newTransformer();
// DOMSource source = new DOMSource(doc);
// StreamResult result = new StreamResult(paper);
// transformer.transform(source, result);
//
// System.out.println("Done");
// }
}
public static Tree HobbsResolve(Tree pronoun, ArrayList<Tree> forest) {
Tree wholetree = forest.get(forest.size()-1); // The last one is the one I am going to start from
ArrayList<Tree> candidates = new ArrayList<Tree>();
List<Tree> path = wholetree.pathNodeToNode(wholetree, pronoun);
System.out.println(path);
// Step 1
Tree ancestor = pronoun.parent(wholetree); // This one locates the NP the pronoun is in, therefore we need one more "parenting" !
// Step 2
ancestor = ancestor.parent(wholetree);
//System.out.println("LABEL: "+pronoun.label().value() + "\n\tVALUE: "+pronoun.firstChild());
while ( !ancestor.label().value().equals("NP") && !ancestor.label().value().equals("S") )
ancestor = ancestor.parent(wholetree);
Tree X = ancestor;
path = X.pathNodeToNode(wholetree, pronoun);
System.out.println(path);
// Step 3
for (Tree relative : X.children()) {
for (Tree candidate : relative) {
if (candidate.contains(pronoun)) break; // I am looking to all the nodes to the LEFT (i.e. coming before) the path leading to X. contain <-> in the path
//System.out.println("LABEL: "+relative.label().value() + "\n\tVALUE: "+relative.firstChild());
if ( (candidate.parent(wholetree) != X) && (candidate.parent(wholetree).label().value().equals("NP") || candidate.parent(wholetree).label().value().equals("S")) )
if (candidate.label().value().equals("NP")) // "Propose as the antecedent any NP node that is encountered which has an NP or S node between it and X"
candidates.add(candidate);
}
}
// Step 9 is a GOTO step 4, hence I will envelope steps 4 to 8 inside a while statement.
while (true) { // It is NOT an infinite loop.
// Step 4
if (X.parent(wholetree) == wholetree) {
for (int q=1 ; q < MAXPREVSENTENCES; ++q) {// I am looking for the prev sentence (hence we start with 1)
if (forest.size()-1 < q) break; // If I don't have it, break
Tree prevTree = forest.get(forest.size()-1-q); // go to previous tree
// Now we look for each S subtree, in order of recency (hence right-to-left, hence opposite order of that of .children() ).
ArrayList<Tree> backlist = new ArrayList<Tree>();
for (Tree child : prevTree.children()) {
for (Tree subtree : child) {
if (subtree.label().value().equals("S")) {
backlist.add(child);
break;
}
}
}
for (int i = backlist.size()-1 ; i >=0 ; --i) {
Tree Treetovisit = backlist.get(i);
for (Tree relative : Treetovisit.children()) {
for (Tree candidate : relative) {
if (candidate.contains(pronoun)) continue; // I am looking to all the nodes to the LEFT (i.e. coming before) the path leading to X. contain <-> in the path
//System.out.println("LABEL: "+relative.label().value() + "\n\tVALUE: "+relative.firstChild());
if (candidate.label().value().equals("NP")) { // "Propose as the antecedent any NP node that you find"
if (!candidates.contains(candidate)) candidates.add(candidate);
}
}
}
}
}
break; // It will always come here eventually
}
// Step 5
ancestor = X.parent(wholetree);
//System.out.println("LABEL: "+pronoun.label().value() + "\n\tVALUE: "+pronoun.firstChild());
while ( !ancestor.label().value().equals("NP") && !ancestor.label().value().equals("S") )
ancestor = ancestor.parent(wholetree);
X = ancestor;
// Step 6
if (X.label().value().equals("NP")) { // If X is an NP
for (Tree child : X.children()) { // Find the nominal nodes that X directly dominates
if (child.label().value().equals("NN") || child.label().value().equals("NNS") || child.label().value().equals("NNP") || child.label().value().equals("NNPS") )
if (! child.contains(pronoun)) candidates.add(X); // If one of them is not in the path between X and the pronoun, add X to the antecedents
}
}
// Step SETTE
for (Tree relative : X.children()) {
for (Tree candidate : relative) {
if (candidate.contains(pronoun)) continue; // I am looking to all the nodes to the LEFT (i.e. coming before) the path leading to X. contain <-> in the path
//System.out.println("LABEL: "+relative.label().value() + "\n\tVALUE: "+relative.firstChild());
if (candidate.label().value().equals("NP")) { // "Propose as the antecedent any NP node that you find"
boolean contains = false;
for (Tree oldercandidate : candidates) {
if (oldercandidate.contains(candidate)) {
contains=true;
break;
}
}
if (!contains) candidates.add(candidate);
}
}
}
// Step 8
if (X.label().value().equals("S")) {
boolean right = false;
// Now we want all branches to the RIGHT of the path pronoun -> X.
for (Tree relative : X.children()) {
if (relative.contains(pronoun)) {
right = true;
continue;
}
if (!right) continue;
for (Tree child : relative) { // Go in but do not go below any NP or S node. Go below the rest
if (child.label().value().equals("NP")) {
candidates.add(child);
break; // not sure if this means avoid going below NP but continuing with the rest of non-NP children. Should be since its DFS.
}
if (child.label().value().equals("S")) break; // Same
}
}
}
}
// Step 9 is a GOTO, so we use a while.
System.out.println(pronoun + ": CHAIN IS " + candidates.toString());
ArrayList<Integer> scores = new ArrayList<Integer>();
for (int j=0; j < candidates.size(); ++j) {
Tree candidate = candidates.get(j);
Tree parent = null;
int parent_index = 0;
for (Tree tree : forest) {
if (tree.contains(candidate)) {
parent = tree;
break;
}
++parent_index;
}
scores.add(0);
if (parent_index == 0)
scores.set(j, scores.get(j)+100); // If in the last sentence, +100 points
scores.set(j, scores.get(j) + syntacticScore(candidate, parent));
if (existentialEmphasis(candidate)) // Example: "There was a dog standing outside"
scores.set(j, scores.get(j)+70);
if (!adverbialEmphasis(candidate, parent))
scores.set(j, scores.get(j)+50);
if (headNounEmphasis(candidate, parent))
scores.set(j, scores.get(j)+80);
int sz = forest.size()-1;
// System.out.println("pronoun in sentence " + sz + "(sz). Candidate in sentence "+parent_index+" (parent_index)");
int dividend = 1;
for (int u=0; u < sz - parent_index; ++u)
dividend *= 2;
//System.out.println("\t"+dividend);
scores.set(j, scores.get(j)/dividend);
System.out.println(candidate + " -> " + scores.get(j) );
}
int max = -1;
int max_index = -1;
for (int i=0; i < scores.size(); ++i) {
if (scores.get(i) > max) {
max_index = i;
max = scores.get(i);
}
}
Tree final_candidate = candidates.get(max_index);
System.out.println("My decision for " + pronoun + " is: " + final_candidate);
// Decide what candidate, with both gender resolution and Lappin and Leass ranking.
Tree pronounparent = pronoun.parent(wholetree).parent(wholetree); // 1 parent gives me the NP of the pronoun
int pos = 0;
for (Tree sibling : pronounparent.children()) {
System.out.println("Sibling "+pos+": " + sibling);
if (sibling.contains(pronoun)) break;
++pos;
}
System.out.println("Before setchild: " + pronounparent);
@SuppressWarnings("unused")
Tree returnval = pronounparent.setChild(pos, final_candidate);
System.out.println("After setchild: " + pronounparent);
return wholetree; // wholetree is already modified, since it contains pronounparent
}
private static int syntacticScore(Tree candidate, Tree root) {
// We will check whether the NP is inside an S (hence it would be a subject)
// a VP (direct object)
// a PP inside a VP (an indirect obj)
Tree parent = candidate;
while (! parent.label().value().equals("S")) {
if (parent.label().value().equals("VP")) return 50; // direct obj
if (parent.label().value().equals("PP")) {
Tree grandparent = parent.parent(root);
while (! grandparent.label().value().equals("S")) {
if (parent.label().value().equals("VP")) // indirect obj is a PP inside a VP
return 40;
parent = grandparent;
grandparent = grandparent.parent(root);
}
}
parent = parent.parent(root);
}
return 80; // If nothing remains, it must be the subject
}
private static boolean existentialEmphasis(Tree candidate) {
// We want to check whether our NP's Dets are "a" or "an".
for (Tree child : candidate) {
if (child.label().value().equals("DT")) {
for (Tree leaf : child) {
if (leaf.value().equals("a")||leaf.value().equals("an")
||leaf.value().equals("A")||leaf.value().equals("An") ) {
//System.out.println("Existential emphasis!");
return true;
}
}
}
}
return false;
}
private static boolean headNounEmphasis(Tree candidate, Tree root) {
Tree parent = candidate.parent(root);
while (! parent.label().value().equals("S")) { // If it is the head NP, it is not contained in another NP (that's exactly how the original algorithm does it)
if (parent.label().value().equals("NP")) return false;
parent = parent.parent(root);
}
return true;
}
private static boolean adverbialEmphasis(Tree candidate, Tree root) { // Like in "Inside the castle, King Arthur was invincible". "Castle" has the adv emph.
Tree parent = candidate;
while (! parent.label().value().equals("S")) {
if (parent.label().value().equals("PP")) {
for (Tree sibling : parent.siblings(root)) {
if ( (sibling.label().value().equals(","))) {
//System.out.println("adv Emph!");
return true;
}
}
}
parent = parent.parent(root);
}
return false;
}
public static ArrayList<Tree> findPronouns(Tree t) {
ArrayList<Tree> pronouns = new ArrayList<Tree>();
if (t.label().value().equals("PRP") && !t.children()[0].label().value().equals("I") && !t.children()[0].label().value().equals("you") && !t.children()[0].label().value().equals("You")) {
pronouns.add(t);
}
else
for (Tree child : t.children())
pronouns.addAll(findPronouns(child));
return pronouns;
}
@SuppressWarnings("rawtypes")
public static ArrayList<List> preprocess(DocumentPreprocessor strarray) {
ArrayList<List> Result = new ArrayList<List>();
for (List<HasWord> sentence : strarray) {
if (!StringUtils.isAsciiPrintable(sentence.toString())) {
continue; // Removing non ASCII printable sentences
}
//string = StringEscapeUtils.escapeJava(string);
//string = string.replaceAll("([^A-Za-z0-9])", "\\s$1");
int nonwords_chars = 0;
int words_chars = 0;
for (HasWord hasword : sentence ) {
String next = hasword.toString();
if ((next.length() > 30)||(next.matches("[^A-Za-z]"))) nonwords_chars += next.length(); // Words too long or non alphabetical will be junk
else words_chars += next.length();
}
if ( (nonwords_chars / (nonwords_chars+words_chars)) > 0.5) // If more than 50% of the string is non-alphabetical, it is going to be junk
continue; // Working on a character-basis because some sentences may contain a single, very long word
if (sentence.size() > 100) {
System.out.println("\tString longer than 100 words!\t" + sentence.toString());
continue;
}
Result.add(sentence);
}
return Result;
}
}
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